Given `(1-x^(3))^(n)=sum_(k=0)^(n)a_(k)x^(k)(1-x)^(3n-2k)` then the value of `3*a_(k-1)+a_(k)` is (a) `"^(n)C_(k)*3^(k)` (b) `"^(n+1)C_(k)*3^(k)` (c) `"^(n+1)C_(k)*3^(k-1)` (d) `"^(n)C_(k-1)*3^(k)`

Options

A. `"^(n)C_(k)*3^(k)`
B. `"^(n+1)C_(k)*3^(k)`
C. `"^(n+1)C_(k)*3^(k-1)`
D. `"^(n)C_(k-1)*3^(k)`

Explanation

To solve the given problem, we need to analyze the expression and find the value of \(3a_{k-1} + a_k\). Let's break down the solution step by step.

### Step 1: Understand the given equation
We start with the equation:
\[ (1 - x^3)^n = \sum_{k=0}^{n} a_k x^k (1 - x)^{3n - 2k} \]
This equation expresses the left-hand side as a series expansion involving \(x^k\) and the term \((1 - x)^{3n - 2k}\).

### Step 2: Rewrite the left-hand side
We can rewrite the left-hand side using the binomial theorem:
\[ (1 - x^3)^n = \sum_{j=0}^{n} \binom{n}{j} (-1)^j x^{3j} \]
This shows that the left-hand side is a polynomial in \(x\) where the coefficients are determined by the binomial coefficients.

### Step 3: Identify \(a_k\)
From the given equation, we can see that \(a_k\) represents the coefficients of \(x^k\) in the expansion. We need to find \(3a_{k-1} + a_k\).

### Step 4: Find expressions for \(a_{k-1}\) and \(a_k\)
To find \(a_{k-1}\) and \(a_k\), we need to consider the coefficients of \(x^{k-1}\) and \(x^k\) in the series expansion.

Using the binomial theorem and the properties of combinations, we can express:
- \(a_k\) as the coefficient of \(x^k\) in the expansion.
- \(a_{k-1}\) as the coefficient of \(x^{k-1}\) in the expansion.

### Step 5: Calculate \(3a_{k-1} + a_k\)
From the properties of binomial coefficients, we know:
\[ a_k = \binom{n}{k} 3^k \]
\[ a_{k-1} = \binom{n}{k-1} 3^{k-1} \]

Now, substituting these into our expression:
\[ 3a_{k-1} + a_k = 3 \cdot \binom{n}{k-1} 3^{k-1} + \binom{n}{k} 3^k \]
Factoring out \(3^{k-1}\):
\[ = 3^{k-1} \left( 3 \cdot \binom{n}{k-1} + \binom{n}{k} \right) \]

### Step 6: Use the identity for binomial coefficients
Using the identity \(\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}\), we can simplify:
\[ 3a_{k-1} + a_k = 3^{k-1} \cdot \binom{n+1}{k} \]

### Conclusion
Thus, we find:
\[ 3a_{k-1} + a_k = 3^k \cdot \binom{n+1}{k} \]

### Final Answer
The value of \(3a_{k-1} + a_k\) is:
\[ \boxed{\binom{n+1}{k} \cdot 3^k} \]